this document presents the Interaction Design Coursework that is used for Greenwich student.
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Soil Structure Interaction below Minor bridge
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HCI
Behavior under Combined Bending and Axial Loads Interaction Diagram Between Axial Load and Moment ( Failure Envelope )
Concrete crushes before steel yields Steel yields before concrete crushes
Note: Any combination of P and M outside the envelope will cause failure.
Example: Axial Load Vs. Moment Interaction Diagram Consider an square column (20 in x 20 in.) with 8 #10 (ρ = 0.0254) and fc = 4 ksi and fy = 60 ksi. Draw the interaction diagram.
Example: Axial Load Vs. Moment Interaction Diagram Point
c (in)
Pn
Mn 0
e
1
-
1451 k
2
17.5
1314 k
351 k-ft
3.2 in
3
12.5
841 k
500 k-ft
7.13 in
4
10.36
585 k
556 k-ft
11.42 in
5
8.0
393 k
531 k-ft
16.20 in
6
6.0
151 k
471 k-ft
37.35 in
7
4.5
0k
395 k-ft
infinity 8
8
0
-610 k
0 k-ft
0
Example: Axial Load Vs. Moment Interaction Diagram Use a series of c values to obtain the Pn verses Mn.
Column Analysis 2000
1400 1500
P (k)
1000 500 0 0
100
200
300
-500 -1000
M (k-ft)
400
500
600
Example: Axial Load Vs. Moment Interaction Diagram Max. compression Column Analysis 1200 1000
Location of the linearly varying φ.
800
φ Pn (k)
600
Cb
400 200 0 -200
0
100
200
300
-400
Max. tension
-600 -800
φ Mn (k-ft)
400
500
Design for Combined Bending and Axial Load (short column) Column Types 3) Tied Column - Bars in 2 faces (furthest from axis of bending. - Most efficient when e/h > 0.2 - rectangular shape increases efficiency
Design for Combined Bending and Axial Load (short column) Spices Typically longitudinal bars spliced just above each floor. (non-seismic) Type of lap splice depends on state of stress (C.12.17)
Design for Combined Bending and Axial Load (short column) Spices All bars in compression
Use compression lap splice (C.12.16)
0 ≤ f s ≤ 0.5 f y on tension face →
fs > 0.5 f y
(< 1/ 2 bars splice) → Class B C.12.15 ( > 1/2 bars spliced) → Class B tension lap splice Class A tension lap
Design for Combined Bending and Axial Load (short column) Column Shear Recall
Nu Vc = 0.17 1 + 14 A g
( Axial Compression )
' λ f bw d c
( C.11-4 )
If Vu > 0.5φVc ⇒ Ties must satisfy C.7.10
Design for Combined Bending and Axial Load (short column) Additional Note on Reinforcement Ratio
Recall 0.01 ≤ ρ ≤ 0.04 (C.10.8 )
∗⇒
( C.10.9.1)
For cross-section larger than required for loading: Min. reinforcement may be computed for reduced effective area, Ag, ( ≥ 1/2 Ag (total) ) Provided strength from reduced area and resulting Ast must be adequate for loading.
Non-dimensional Interaction Diagrams Pn f c Ag
versus
Pn or K n = f c Ag
Mn f c Ag h
versus
Pn e Rn = f c Ag h
See Figures B-12 to B-26 or ACI Common 340 Design Handbook Vol 2 Columns (ACI 340.2R-91)
Non-dimensional Interaction Diagrams
Design using non-dimensional interaction diagrams 1.) Calculate factored loads (Pu , Mu ) and e for relevant load combinations 2.) Select potentially governing case(s) 3.) Use estimate h to calculate γh, e/h for governing case(s)
Design using non-dimensional interaction diagrams 4.) Use appropriate chart (App. A) target ρg Pn f c Ag
Pf ⇒ Calculate Ag = u c required φ Pn f c Ag (for each governing case)
Read
5.) Select b & h ⇒ Ag = b * h
Design using non-dimensional interaction diagrams 6.) If dimensions are significantly different from estimated (step 3), recalculate ( e / h ) and redo steps 4 & 5. Revise Ag if necessary. 7.) Select steel
⇒
Ast = ρ Ag
Design using non-dimensional interaction diagrams 8.) Using actual dimensions & bar sizes to check all load combinations ( use charts or “exact: interaction diagram).
9.) Design lateral reinforcement.
Example: Column design using Interaction Diagrams Determine the tension and compression reinforcement for a 16 in x 24 in. rectangular tied column to support Pu= 840 k and Mu = 420 k-ft. Use fc = 4 ksi and fy = 60 ksi. Using the interaction diagram.
Example: Interaction Diagrams Compute the initial components Pu
840 kips = = 1292 k Pn = φ 0.65 12 in. 420 k-ft Mu ft = 6.0 in. en = = Pu 840 k
Example: Interaction Diagrams Compute the initial components
γ h = 24 in. − 5.0 in. = 19.0 in. 19.0 in. γ= = 0.79 24 in.
Example: Interaction Diagrams Compute the coefficients of the column
Pn 1292 k Kn = = Ag f c (16 in.)( 24 in.)( 4 ksi ) = 0.84 1292 k )( 6 in.) ( Pn e Rn = = Ag f c h (16 in.)( 24 in.)( 4 ksi )( 24 in.) = 0.21
Example: Interaction Diagrams Using an interaction diagram, B-13
( Rn , K n ) = ( 0.21, 0.84 ) γ = 0.7 f c = 4 ksi f y = 60 ksi
ρ = 0.042
Example: Interaction Diagrams Using an interaction diagram, B-14
( Rn , K n ) = ( 0.21, 0.84 ) γ = 0.9 f c = 4 ksi f y = 60 ksi
ρ = 0.034
Example: Interaction Diagrams Using linear interpolation to find the ρ of the column
Example: Interaction Diagrams The second equation comes from the equilibrium equation and substitute in for Pn
Pn = Cs1 + Cc − T 541.2 k + 64.14c − 1.27c = 441.5 k + 46.24c − 7.8 fs 2
7.8 f s = 1.27c − 17.9c − 99.7 2
f s = 0.1628c − 2.282c − 12.782 2
Example: Interaction Diagrams Substitute the relationship of c for the stress in the steel. 21.5 in. − c 2 = c − 2.282c − 12.782 87 0.1628 c The problem is now a cubic solution c 15 in. 19 in. 19.5 in. 20.0 in. 19.98 in.
fs 37.7 11.45 8.92 6.52 6.62
RHS -10.38 2.64 4.63 6.70 6.62
Example: Interaction Diagrams Compute Pn
Pn = 541.2 k + 64.14 (19.98 in.) − 1.27 (19.98 in.) = 1313.7 k > 1292 k Compute Mn about the center
h a h h M n = Cs1 − d ′ + Cc − + T d − 2 2 2 2
2
Example: Interaction Diagrams Compute Mn about the center
M n = 441.5 k (12 in. − 2.5 in.) 0.85 (19.98 in.) +46.24 (19.98 in.) 12 in. − 2 + ( 7.8 in